Introduction to Thermodynamics

Rub your hands together for 15 seconds.

Are your hands warm?

Thermal energy

The study of the effects of work, heat flow, and energy on a system

Movement of thermal energy

Engineers use thermodynamics in systems ranging from nuclear power plants to electrical components.

SYSTEM

SURROUNDINGS

BOUNDARY

Thermal Energy is kinetic energy in transit from one object to another due to temperature difference. (Joules)

Temperature is the average kinetic energy of particles in an object – not the total amount of kinetic energy particles. (Degrees)

Temperature #1 Temperature #2

Heat

Scale Freezing point of water

Boiling point of water

Celsius 0°C 100°C

Fahrenheit 32°F 212°F

Kelvin 273K 373K

Matter is made up of molecules in motion (kinetic energy)

An increase in temperature increases motion

A decrease in temperature decreases motion

Absolute Zero occurs when all kinetic energy is removed from a object 0 K = -273° C and o F = -460° R

Thermal equilibrium is obtained when touching objects within a system reach the same temperature.

This occurs through Thermal Energy Transfer, i.e (literally) the movement or transfer of thermal energy. Commonly occurs via: Convection Conduction Radiation

When thermal equilibrium is reached, the system loses its ability to do work. Why is this?

When thermal equilibrium is reached, the system loses its ability to do work. Why is this?

To answer, an advanced concept needs to be introduced: the gradient

A gradient can have many forms:Mathematical

HeightEnergy

Thermodynamics deals with energy gradients

So what is it?

In general, a gradient is the imbalance created by a difference in something.

A gradient can have many forms:Mathematical (5, 4, 3, 2, 1…)

Height (Cliff, Waterfall, Hil…)Energy (Battery, Fire, Falling weight…)Sounds simple right?

Remember: Thermo deals with energy gradients

Govern the transfer, exchange, conversion, etc. of thermal and other forms of energy

Consequences include:

100% efficiency is unattainable

All Energy conversion processes are irreversible

Perpetual motion and spontaneous organization impossible

Thermodynamics has the most profound consequences of any of the laws of science

Zeroth Law of Thermodynamics:

If two systems are separately found to be in thermal equilibrium with a third system, the first two systems are in thermal equilibrium with each other.

Thermodynamic Laws

Law of energy conservation applied to a thermal system Energy cannot be created or destroyed.

Energy can change form and location.

The total amount of energy remains constant.

Note: Thermal energy can be increased within a system by adding thermal energy or by performing work on a system, but the total amount of energy in existence doesn’t increase.

Example: Using a bicycle pump

Pumping the handle results in what?Adding mechanical energy to the system

Some mechanical energy is converted into kinetic energy (the pressure increases)

Some mechanical energy is converted into thermal energy (the pump becomes hot)

The total increase in internal energy of the system is equal to what?The applied mechanical energy

No new energy is created

©iStockphoto.com

An energy balance or an accounting statement of the quantity of energy.

Sum of energyinputs

Sum of energyoutputs

Net changeof energyin system

Specifies the direction that a certain energy transformation can follow:Heat energy flows down a gradient naturally; the reverse is not possible without additional input of energy.

Thermal energy flows from hot to cold

When you touch a cooked pizza with your hand, thermal energy flows in what direction?

When you touch a frozen pizza with your hand, thermal energy flows in what direction?

Hand → Pizza

Pizza → Hand©iStockphoto.com

©iStockphoto.com

Entropy is the measure of how evenly distributed heat is within a system.

- a system tends to go from order to disorder

OrderOrder DisorderDisorder

Firewood has low entropy (molecules in order) when stacked and high entropy when burning (molecules in disorder).

The total amount of energy in the world does not change, but the availability of that energy constantly decreases.

It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W . Some amount of heat QC must be exhausted to a cold reservoir. This precludes a perfect heat engine.

• Second Law of Thermodynamics: In any cyclic process the entropy will either increase or remain the same.

Entropy: a measure of the amount of energy which is unavailable to do work.

Entropy: a measure of the disorder of a system.

• •Since entropy gives information about the evolution of an isolated system with time, it is said to give us the direction of "time's arrow" .

The First Law of Thermodynamics says you can't get something for nothing…

The Second Law of Thermodynamics says you can't even break even.

ConvectionThe transfer of thermal energy by movement of fluid (liquid or gas)

When fluid is heated, it expands, becomes less dense, and rises.

Boiler heating systems circulate heat throughout a home without pumps through the use of convection.

Conduction The transfer of thermal energy within an object or between objects from molecule to molecule

A metal spoon placed in a hot cup of soup will feel warm to your hand. The heat from the soup is conducted through the spoon. ©iStockphoto.com

pQ = m C ΔT

p

Q = energy transfer (Joules)

m = mass of the material (kilograms)

C =specific heat capacity of the material (J / kg C)

T = temperature

Δ = difference

QP =

Δt

ΔTP = kA

L

P = rate of energy transfer (Watts)

Q = energy transfer (Joules)

t = time(seconds)

k = thermal conductivity

A = area of thermal conductivity

L = thickness of material

T = temperature

Δ = difference

PLk =

AΔT

Step 1. List all known values

ΔT = 30.0°C – 25.0°C = 5.0°C

4184 o

J

kg× C

1 kg

Q

Calculate the energy transferred when a block of aluminum at 80.0°C is placed in 1.00 liter (1kg) of water at 25.0°C if the final temperature becomes 30.0°C.

Mass of water =

Cp of water =

Difference in temperature =

Cp of Al = 900. o

J

kg× C

ΔT = 80.0°C – 30.0°C = 50.0°C Difference in temperature =

Step 2. List all unknown values

Q = energy transferredmAl = mass of the Al block

Step 3. Select equations to solve unknown values

pQ = m C ΔT Al waterQ = Q

Step 4. Solve for Qwater

owater o

JQ = (1.00kg) •4184 5.0 C = 21,000 J gained

kg× C

Step 5. Solve for mAl

Al waterQ (lost) = Q (gained) = 20,920 J

AlAl

opo

Q 20,920Jm = =

JC ΔT 900. 50.0 Ckg C

Alm = 0.46 kg = 460g

Al AlQ = m Cp ΔT

Step 1. List all known values

ΔT = 10°C - 5°C = 5°C

Δt = 1 hour = 3600s

k =0.10 o

J

s×m× C

L = 0.04m

A = 2m * 1m = 2m2

Q

Calculate the energy transfer in a wall section measuring 2m by 1m by 0.04m thick

with a thermal conductivity of 0.10 . Opposing sides of the wall section have a temperature of 10°C and 5°C after one hour.

J

s×m×°C

Area of thermal conductivity =

Thermal conductivity =

Thickness of material =

Difference in temperature =

Difference in Time =

Step 2. List all unknown values

P = Rate of energy transferQ = Energy transfer

Step 3. Select equations to solve unknown values

QP =

Δt

ΔTP = kA

L

Step 4. Solve in terms of Q

Q = P • Δt

Step 5. Combine equations

ΔTQ = (kA ) • Δt

L

Step 6. Apply known values

ΔTQ = (kA ) • Δt

L

2J 5°C0.10 • 2m • •3600s

s×m×°C 0.04m

Q

Q = 90,000J

Overall heat coefficient

The measure of a material’s ability to conduct heat

2

Btu

ft hr °F

2

W

m °C

Thermal Conductivity of a Material

U.S. customary system

Metric system

PU =

AΔT

R-Value

The measure of a material’s ability to resist heat

The higher the R-value, the higher the resistance

Bulk R-value = R-value Object 1 + R-value Object 2 + … = Total R-Value

Thermal Resistance of a Material

1R =

U

PU =

AΔT

Determine the R-value of the wall cavity below

5/8 in. drywall(R=0.56)

Fiberglass batt(R=19)

1 in. foil-faced polyisocyanurate(R=7.20)

1 in. air space(R=0.17)

Brick 2 ¼ x 3 ½ x 8(R=0.8)

2x6 construction(2x6 R=6.88)

0.56+19.00+ 7.20+ 0.17+ 0.8

= 27.7

Wall cavity R-value What is the R-value at a stud location?

0.56+ 6.88+ 7.20+ 0.17+ 0.8=15.6

RadiationThe process by which energy is transmitted through a medium, including empty space, as electromagnetic waves

e emissivity constant

T temperature in Kelvin

Electromagnetic waves transfer to food and other matter

All objects lose and gain thermal energy by electromagnetic radiation.

4 4net 2 1P = σAe(T - T )

-82 4

P =radiated energy transfer

Wσ =Stefan's constant = 5.6696x10

m ×KA =area

Stefan’s Law

©iStockphoto.com

Area

Emissivity constant

Stefan’s constant

Bedroom temperature

Skin temperature

Time in seconds

21.= = 30mA

=e=0.90-8

2 4

W5.6696 •10=

m=

× Kσ

1=T =65°F

Prior to dressing for school, a student watches the morning weather to decide what clothes to wear. The bedroom is 65ºF and the student’s skin is 91.4ºF. Determine the net energy transfer from the student’s body during the 15.0 minutes spent watching the morning weather. Note: Skin emissivity is 0.90, and the surface area of the student is 1.30m2.

Step 1. List all known values

= = 15.0 minutes =t 900.s

2=T =91.4°F

©iStockphoto.com

4 4net 2 1P = σAe(T - T )

2 14 4 ( T -T )= 8,784,904,710.59K - 7,218,301,921.12K

Step 2. List all unknown values

P = Rate of energy transferQ = Energy transfer

Step 3. Select equations to solve unknown values

Step 4. Apply known values to 2 14 4

netP =σAe( T -T )

Q = P • Δt

2 14 4Solve for( T -T )

91.4°F =306.15K

65°F = 291.48K

2 14 4 ( T -T )= 1566602759.47K

©iStockphoto.com

-8 2net 2 4

WP = 5.6696 10 1.30m 0.90 1,566,602,759.47K

m ×K

Step 4 (continued). Apply known values to 4 4net 2 1P = σAe T - T

netP =103.92W

Q = PΔt

Step 5. Combine equations and solve

Q =103.92W 900.s

Q =93,500J

93,500J of energy are transferred from the student’s body during the 15 minutes spent watching the morning weather.

©iStockphoto.com

http://www.nrel.gov

All images were obtained from the following URL: http://www1.eere.energy.gov

Energy generated from the thermal energy stored beneath the Earth’s surface

Also refers to the heat that is collected from the atmosphere; for instance, near the oceans

Energy Information Association. (n.d.). Energy kid’s page. Retrieved March 23, 2008, from http://www.eia.doe.gov/kids/energyfacts

McGraw-Hill dictionary of engineering. (2nd ed.). New York, NY: McGraw-Hill.

NASA. (2008). Glenn research center. Retrieved March 23, 2008 from http://www.nasa.gov/centers/glenn/home/index.html

National Renewable Energy Laboratory. (2007). TroughNet. Retrieved March 23, 2008, from http://www.nrel.gov/csp/troughnet

U.S. Department of Energy. (2008). Solar energy technologies program. Retrieved March 23, 2008, from http://www1.eere.energy.gov/solar